S2 June 2013 (R) Q4

EdexcelOld spec11 marksContinuous Random Variables

4. The random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by

\[\mathrm{f}(x) = \begin{cases} k(3 + 2x - x^2) & 0 \leqslant x \leqslant 3 \\ 0 & \text{otherwise} \end{cases}\]

where \(k\) is a constant.

(a) Show that \(k = \dfrac{1}{9}\) (3)

(b) Find the mode of \(X\). (2)

By comparing your answers to parts (b) and (c),

(d) describe the skewness of \(X\), giving a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int \mathrm{f}(x)\,\mathrm{d}x = k\left[3x + x^2 - \frac{x^3}{3}\right]\)	M1
\(\displaystyle\int_0^3 \mathrm{f}(x)\,\mathrm{d}x = 1\) gives \(k\left[\left(9 + 9 - \dfrac{27}{3}\right) - (0)\right] = 1\)	M1
So \(k = \dfrac{1}{9}\) (*)	A1cso
	(3)

Notes

NB This is a ‘Show that so working must be seen’

1st M1 for some correct integration \(x^n \to x^{n+1}\) for at least one term

2nd M1 for some correct use of the limit 3 and at least implied use of limit 0 and put =1

A1cso for correct solution with no incorrect working seen.

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}'(x) = k(2 - 2x)\)	M1
\(\mathrm{f}'(x) = 0\) implies \(x = 1\) so mode = 1	A1
	(2)

Notes

M1 for attempt to differentiate and putting = 0. At least one correctly differentiated \(x\) term. or for an alternative method for finding the maximum such as completing the square and selecting the corresponding \(x\) value or using a sketch and symmetry.

A1 for mode = 1

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(X) = \displaystyle\int_0^3 \tfrac{1}{9}\left(3x + 2x^2 - x^3\right)\,\mathrm{d}x\)	M1
\(= \dfrac{1}{9}\left[\dfrac{3x^2}{2} + \dfrac{2x^3}{3} - \dfrac{x^4}{4}\right]_0^3\)	M1dA1
\(= \left\{\dfrac{1}{9}\left[\left(\dfrac{3}{2} \times 9 + \dfrac{2}{3} \times 27 - \dfrac{81}{4}\right) - 0\right]\right\} = \dfrac{5}{4}\)	A1
	(4)

Notes

1st M1 for clear attempt to use \(x\mathrm{f}(x)\) with an intention of integrating (Integral sign enough) Ignore limits. Must substitute in \(\mathrm{f}(x)\)

2nd M1d dependent on 1st M being awarded. For some correct integration...at least one correct term with the correct coefficient.

1st A1 for fully correct (possibly un-simplified) integration. Ignore limits

2nd A1 for answer of 5/4 or 1.25 or some other exact equivalent

Mark scheme (d)

Scheme	Marks
Mean > mode	M1
So positive skew	A1
	(2)
	(11 marks)

Notes

M1 for a comparison of mean and mode (ft their values of mode and mean). Do not allow median.

A1 for positive skew only (provided this is compatible with their values and comparison)